Problem 68 - Intermediate Algebra

Question

Rationalize each denominator. All variables represent positive real numbers. $$ \frac{\sqrt{15 b^{2}}}{\sqrt{5 b^{3}}} $$

Step-by-Step Solution

Answer

The rationalized expression is \( \frac{\sqrt{3b}}{b} \).

Step 1: Simplify under the radical

We begin by simplifying the expression under the square roots. The numerator is already simplified, being \( \sqrt{15b^2} \). In the denominator, \( \sqrt{5b^3} \) can be split into \( \sqrt{5} \times \sqrt{b^3} \). Simplifying \( \sqrt{b^3} \) gives \( b^{3/2} \) or \( b\sqrt{b} \). Therefore, the denominator becomes \( \sqrt{5}b\sqrt{b} \).

Step 2: Simplify the whole fraction

Now express the original fraction as \( \frac{\sqrt{15b^2}}{\sqrt{5}b\sqrt{b}} \). Simplifying \( \sqrt{15b^2} \) results in \( b\sqrt{15} \). Therefore, the fraction becomes \( \frac{b\sqrt{15}}{\sqrt{5}b\sqrt{b}} \), which reduces to \( \frac{\sqrt{15}}{\sqrt{5}\sqrt{b}} \) since \( b \) cancels out.

Step 3: Rationalize the denominator

To rationalize this denominator, we multiply numerator and denominator by \( \sqrt{5b} \). The denominator becomes \( \sqrt{5} \times \sqrt{5b} = 5 \times b \). The numerator becomes \( \sqrt{15} \times \sqrt{5b} = \sqrt{75b} = \sqrt{25 \times 3b} = 5\sqrt{3b} \).

Step 4: Write the final simplified expression

Thus, the expression becomes \( \frac{5\sqrt{3b}}{5b} \). Simplifying this gives \( \frac{\sqrt{3b}}{b} \).

Key Concepts

Simplifying Radicals

Simplifying radicals involves reducing an expression under a square root to its simplest form. This is crucial when working with algebraic fractions, as it makes calculations more manageable. Let's dive into the process of simplifying radicals:

• First, identify the components under the radical sign. For instance, with expressions like \( \sqrt{15b^2} \), recognize that part of the expression can be simplified because \( b^2 \) is a perfect square, resulting in \( b\sqrt{15} \).
• If the expression involves multiple terms, try to factor each term to identify perfect squares. For example, with \( \sqrt{5b^3} \), we can break it down to \( \sqrt{5} \times b\sqrt{b} \), since \( b^3 = b^2 \times b \).

Understanding these steps helps you streamline expressions for easier manipulation in equations and further algebraic processes.

Square Roots

Square roots are at the heart of simplifying expressions involving radicals. Grasping square roots is essential because they help in converting complex expressions into more understandable forms.
Here's how we handle square roots:

• Know that the square root of a number \( x \), represented as \( \sqrt{x} \), is a value that gives you \( x \) when multiplied by itself. For example, \( \sqrt{16} = 4 \) because \( 4 \times 4 = 16 \).
• Square roots act as the reverse operation of squaring; therefore, simplifying square roots means finding any perfect squares within a product. In the exercise \( \sqrt{75b} \), notice \( 75 = 25 \times 3 \), thus \( \sqrt{75} = 5\sqrt{3} \).

The ability to identify and simplify square roots facilitates solving complex algebra problems by making fractions and expressions more manageable.

Algebraic Fractions

Algebraic fractions involve fractions where the numerator and/or denominator contain algebraic expressions. These can become quite complex if not simplified appropriately. Let’s take a closer look at how to manage them effectively:

• Begin by simplifying each part of the fraction separately. Simplify the numerator and denominator independently, working out any radicals or polynomial terms.
• Look out for opportunities to cancel out common terms in the numerator and denominator. For example, in \( \frac{b\sqrt{15}}{\sqrt{5}b\sqrt{b}} \), you can cancel \( b \) from the numerator and denominator, streamlining the fraction.
• Once simplified, consider rationalizing the fraction to eliminate any radicals in the denominator; in our exercise, this is done by multiplying and conjugating to remove the square root from the denominator.

Understanding algebraic fractions and their simplification methods makes solving equations and expressions easier and more efficient.